How can you simplify the quadratic formula using completing the square?

[Anamitra]

Post #77 has been overlooked by micromass and Pwsnafu in a host of other commitments.
Link:

\https://www.physicsforums.com/showpost.php?p=3401056&postcount=77

 

[micromass]

$${i}^{4}{=}{1}$$

$${{(}{i}^{4}{)}}^{1/2}{=}{1}^{1/2}$$

$${i}^{2}{=}{+}{1}$$

The "no exceptions" clause disallows any prescription/rule[ an exception handling clause to] handle this situation

Well, that just illustrates that $(a^b)^c=a^{bc}$ isn't necessarily true for complex numbers

 

[Anamitra]

$${(}{{i}^{4}{)}}^{1/2}{=}{(}{\sqrt{i}{)}}^{4}$$

You agree to this[Your rule does not suggest this unless you incorporate some exception handling clause]?

 

[micromass]

$${i}^{4}$$ is not a complex number

It's not a complex number, so... ?

 

[agentredlum]

The square root symbol is only defined for positive real numbers. So writing $\sqrt{-1}$ isn't defined. (Yes, I know that there are math books out there that do use the notation $i=\sqrt{-1}$, but I still don't consider that notation to be standard).

You have to define sqrt(-1) = i otherwise you lose Gauss Fundamental Theorem of Algebra...and nobody wants to lose THAT.

 

[micromass]

You have to define sqrt(-1) = i otherwise you lose Gauss Fundamental Theorem of Algebra...and nobody wants to lose THAT.

No, you have to define $i^2=1$ for that. That is not the same as saying $\sqrt{-1}=i$. The square root operator is only defined for positive real numbers! But just because we didn't define $\sqrt{-1}$ as -1, doesn't mean that $i^2=1$ isn't true!

 

[agentredlum]

No, you have to define $i^2=1$ for that. That is not the same as saying $\sqrt{-1}=i$. The square root operator is only defined for positive real numbers! But just because we didn't define $\sqrt{-1}$ as -1, doesn't mean that $i^2=1$ isn't true!

Then how would you solve x^2 = -1

By inspection? By factoring?

The extraction of roots must be made legitimate over all finite polynomials with integer co-efficients.

i^2 = -1 does not do this

sqrt(-1) = i does.

 

[micromass]

Then how would you solve x^2 = -1

By inspection? By factoring?

The extraction of roots must be made legitimate over all finite polynomials with integer co-efficients.

i^2 = -1 does not do this

sqrt(-1) = i does.

Well, how would you solve $x^2=2+i$?? By saying $x=\sqrt{2+i}$?? How helpful...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of $x^n=-1$:

$$\cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}$$

but by your method, you would only find $x=\sqrt[n]{x}$ (even if we would allow the square root on complex numbers). That's not really what we want, is it??

 

[agentredlum]

Well, how would you solve $x^2=2+i$?? By saying $x=\sqrt{2+i}$?? How helpful...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of $x^n=-1$:

$$\cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}$$

but by your method, you would only find $x=\sqrt[n]{x}$ (even if we would allow the square root on complex numbers). That's not really what we want, is it??

micromass asked...

"Well, how would you solve $x^2=2+i$?? By saying $x=\sqrt{2+i}$?? How helpful..."

I would throw caution to the wind and let the chips fall where they may...THEN...

I would put this particular example into the quadratic formula and then sustitute back to confirm i got the right answer. Messere Gauss proved I can do it this way.

If you post a more complicated example which does not have a closed form solution i can still get close to the answer by using the 4 binary operations AND THE EXTRACTION OF ROOTS.

The ability to extract roots of real numbers is necessary, this means you need to extract roots of negative values.

The Real numbers are 1 dimensional, Complex numbers NOT on the vertical axis have 2 distinct dimensions.

How fortunate that i is on the vertical axis and has length 1? So we can use it as a 'cheat' to simplify the algebra.

You are correct as far as the cyclotomic x^n = a + bi is concerned

However using sines and cosines to get the answer of course produces multi values since they are periodic!

This is a neat trick, and i have solved x^n = a + bi in Linear Algebra so i know what you mean and agree with you on most of your points.

micromass also said...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of $x^n=-1$:

To the best of my knowledge this method you describe works only if you don't have a MIDDLE TERM.

x^2 = (1 + 2i)x + i would not work with this method but easily works using quadratic formula.

 

[pwsnafu]

You have to define sqrt(-1) = i otherwise you lose Gauss Fundamental Theorem of Algebra...and nobody wants to lose THAT.

See below.

If you post a more complicated example which does not have a closed form solution i can still get close to the answer by using the 4 binary operations AND THE EXTRACTION OF ROOTS.

x^5 - x + 1 = 0

No idea what you mean by "close" or "extraction of roots".

The ability to extract roots of real numbers is necessary,

What? Why?

this means you need to extract roots of negative values.

In real analysis? Hell no!

The Real numbers are 1 dimensional, Complex numbers NOT on the vertical axis have 2 distinct dimensions.

How fortunate that i is on the vertical axis and has length 1? So we can use it as a 'cheat' to simplify the algebra.

The complex numbers are defined as R[X]/(X²+1) and i is defined as the cosets of X. That's the definition! What is this talk of "cheat" or "square roots"?

Please answer the following: how much complex analysis have you done? Because you are arguing about stuff they teach you in the first week.

You are correct as far as the cyclotomic x^n = a + bi is concerned

However using sines and cosines to get the answer of course produces multi values since they are periodic!

Here is micromass's result: $$\cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}$$. Did you notice that k = 0, 1, ..., n-1 and stops? It's a finite number of solutions. Specifically, you get n, agreeing with the fundamental theorem of algebra.

micromass also said...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of $x^n=-1$:

To the best of my knowledge this method you describe works only if you don't have a MIDDLE TERM.

x^2 = (1 + 2i)x + i would not work with this method but easily works using quadratic formula.

And again solve x^5 - x + 1 = 0.

 

[agentredlum]

See below.
x^5 - x + 1 = 0

No idea what you mean by "close" or "extraction of roots".
What? Why?
In real analysis? Hell no!
The complex numbers are defined as R[X]/(X²+1) and i is defined as the cosets of X. That's the definition! What is this talk of "cheat" or "square roots"?

Please answer the following: how much complex analysis have you done? Because you are arguing about stuff they teach you in the first week.
Here is micromass's result: $$\cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}$$. Did you notice that k = 0, 1, ..., n-1 and stops? It's a finite number of solutions. Specifically, you get n, agreeing with the fundamental theorem of algebra.
And again solve x^5 - x + 1 = 0.

Look at this link, what are all those radicals doing there?

http://mathworld.wolfram.com/QuinticEquation.html

What do you know about closure?

Are you saying we don't need radicals to approximate solutions to polynomial equations?

Why are the Complex numbers closed under addition subtraction multiplication division AND EXTRACTION OF ROOTS?

 

[agentredlum]

Why stop at n-1 why not keep going?

You get the same solutions with different numbers inside the cosine and sine.

You 'sound' like you know what you are talking about, throwing around fancy terms...Real Analysis...big deal, but you only see it your way.

I GAURANTEE you that your calculator does not use Real Analysis and cosets to approximate roots of polynomial equations.

1669Newton introduces his iterative method for the numerical approximation of roots.

 

[agentredlum]

Where did you come from? micromass does not need you to defend him/her, he/she is perfectly capable of holding a rational, intelligent stimulating conversation so just chill.

 

[agentredlum]

How does solving x^5 = x - 1 relate to what i said to micromass about the middle term?

Can you factor x^4 + x^2 + 1

??

 

[pwsnafu]

Look at this link, what are all those radicals doing there?
http://mathworld.wolfram.com/QuinticEquation.html

How does solving x^5 = x - 1 relate to what i said to micromass about the middle term?

Can you factor x^4 + x^2 + 1

??

I'm putting these two together. You made the claim that "I would put this particular example into the quadratic formula and then sustitute back to confirm i got the right answer. Messere Gauss proved I can do it this way." Which is fine. But the http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem" says you can't do this for a general quintic. Specifically the example I gave you cannot be solved using radicals. It's got nothing to do with middle term claim you made or approximations.

What do you know about closure?

That fields are closed under their operations? That the complex numbers are algebraically closed? What do you want to know?

Are you saying we don't need radicals to approximate solutions to polynomial equations?

Well, if you only care about Newton's method, then that method doesn't use nth roots. Only subtraction, division and derivatives.

Why are the real numbers closed under addition subtraction multiplication division AND EXTRACTION OF ROOTS?

Well add/sub/mult/div I'm not going into. There are plenty of books on that. I'm interpreting "extraction of roots" to mean real exponentiation. It is is closed in the sense that a^b is always a real number. But it is only defined when

1. a is non-negative and b is real,
2. b is an odd integer and a is real.

So it's not a binary operation over all the reals. It is closed in the sense that its codomain is still the reals.

Why stop at n-1 why not keep going?

You get the same solutions with different numbers inside the cosine and sine.

Correct that is why you stop. You don't need the rest, they just cycle around. It's where the term "cyclotomic" comes from.

You 'sound' like you know what you are talking about, throwing around fancy terms...Real Analysis...big deal, but you only see it your way.

Borderline ad hominem. Fancy or not, they are the mathematical terms. And this is a mathematical forum.

I GAURANTEE you that your calculator does not use Real Analysis and cosets to approximate roots of polynomial equations.

1669Newton introduces his iterative method for the numerical approximation of roots.

Considering Newton's method uses the derivative, actually it does.
Sometimes Newton's Method results in cycles, and it's useful to know when that happens. Although most of the advanced calculators such as Mathematica have probably moved over to the superior http://en.wikipedia.org/wiki/Jenkins-Traub_method" . I don't know any root finding algorithms that don't use the derivative at all.

Where did you come from? micromass does not need you to defend him/her, he/she is perfectly capable of holding a rational, intelligent stimulating conversation so just chill.

Post number 15.

 

[agentredlum]

And if your function contains a radical what are the derivatives going to give you? MORE RADICALS!

When i say radical i mean an nth root or a power (NOT n) of an nth root.

Look i gave an example where i believe method mentioned by micromass fails but my method works.

You gave an example where both methods fail, so you really didn't help his case.

I can't factor x^5 - x + 1

But i can factor x^4 + x^2 + 1 so i challenge you to write this as a product of 2 expressions without using complex numbers

You got the theorem you mention 99% correct. Them 2 proved you cannot solve certain quintics using a FINITE number of operations, but you can approximate the roots of any polynomial to any degree of accuracy using infinite series, or other methods started by Vieta and continued by Cauchy, Weirstrauss, Tsirnhause, Klien and many others too numerous to list.

 

[agentredlum]

Oh shucks...i just looked at post 15. THAT IS A GREAT TRICK!

I hope i have not offended you, i guess i got a little defensive.

You didn't send a picture so i have to use my BRAIN to decode TeX because my browser does not decode it.

In post 15 you prove an irrationl number to an irrational power can be rational and indeed an integer in that example! (If sqrt(2)^sqrt(2) is irrational then you can raise it to an irrational power and turn it into 2. Either way your claim is confirmed)

Well done!

PLEASE POST MORE TRICKS!

 

[dodo]

I wonder why do you insist on your browser "not decoding TeX". The formulas are turned into images on the server side; there is no such thing as an "incapable browser" in this sense.

If you are seeing broken or repeated images, you may want to try clearing the browser's cache and then reloading the page. Hope this helps!

 

[agentredlum]

x^5 - x + 1 = 0 cannot be solved using a finite number of radicals but it can be solved using an INFINITE number of radicals. I remember reading about this in a number theory book. If someone can confirm or deny I would be greatfull.

 

[agentredlum]

I wonder why do you insist on your browser "not decoding TeX". The formulas are turned into images on the server side; there is no such thing as an "incapable browser" in this sense.

If you are seeing broken or repeated images, you may want to try clearing the browser's cache and then reloading the page. Hope this helps!

I am using Playstation 3, not a computer and it has limitations. For another example, i can't read pdf files.

 

[pwsnafu]

And if your function contains a radical what are the derivatives going to give you? MORE RADICALS!

When i say radical i mean an nth root or a power (NOT n) of an nth root.

You said polynomial previously. Are you talking about irrational coefficients?

Look i gave an example where i believe method mentioned by micromass fails but my method works.

You gave an example where both methods fail, so you really didn't help his case.

This threadjack started because of misapplication of square roots, specifically trying to apply (a^b)^c = a^bc to complex numbers, waaaay back in post #64. And it exploded from there.

I can't factor x^5 - x + 1

But i can factor x^4 + x^2 + 1 so i challenge you to write this as a product of 2 expressions without using complex numbers

Do you mean x⁴+x²+1 = (x²+x+1)(x²-x+1)?
You do this kind of thing in ring theory. We say "show x⁴+x²+1 is not irreducible on Z[X]". Standard homework question.

You got the theorem you mention 99% correct. Them 2 proved you cannot solve certain quintics using a FINITE number of operations, but you can approximate the roots of any polynomial to any degree of accuracy using infinite series, or other methods started by Vieta and continued by Cauchy, Weirstrauss, Tsirnhause, Klien and many others too numerous to list.

Yup. Technically still wrong. You can't express certain roots using finite number of addition, subtraction, multiplication, division and nth root. If you include more operations, you're fine.

Anyway, infinite series and approximation techniques is what analysis is about. So you are doing real analysis. Kinda my point.

 

[agentredlum]

You said polynomial previously. Are you talking about irrational coefficients?
This threadjack started because of misapplication of square roots, specifically trying to apply (a^b)^c = a^bc to complex numbers, waaaay back in post #64. And it exploded from there.
Do you mean x⁴+x²+1 = (x²+x+1)(x²-x+1)?
You do this kind of thing in ring theory. We say "show x⁴+x²+1 is not irreducible on Z[X]". Standard homework question.
Yup. Technically still wrong. You can't express certain roots using finite number of addition, subtraction, multiplication, division and nth root. If you include more operations, you're fine.

Anyway, infinite series and approximation techniques is what analysis is about. So you are doing real analysis. Kinda my point.

Yes! I am pleasantly surprised at how quickly you factored x^4 + x^2 + 1

This is a thread about math tricks (see title) and i know a trick to factoring x^4 + x^2 + 1

step 1 Add zero to x^4 + x^2 + 1 in the form (x - x) and re-arrange these 2 parts that equal zero in a clever way.

x^4 + x + x^2 - x + 1

step 2 Group the first 2 terms and the last 3 terms

(x^4 + x) + (x^2 - x + 1)

step 3 factor the first group

x(x^3 + 1) + (x^2 - x + 1)

x(x + 1)(x^2 - x + 1) + (x^2 - x + 1)

now you have a common factor of (x^2 - x + 1) I have used the sum of two cubes factoring technique,

(x^2 - x + 1)[x(x + 1) + 1]

step 4 simplify inside brackets

(x^2 - x + 1)(x^2 + x + 1)

There it is, in my opinion almost like magic, neat huh?

P.S. Only rational coefficients and i see your trap about the derivative, very clever sir. Whats the point of arguing further unless i am VERY mistaken. Let's be friends and post more tricks.

 

[agentredlum]

Anyone with some algebra skill can use (x - x) to derive factoring sum of two cubes formula and diference of two cubes formula.

I INVITE YOU ALL TO TRY!

P.S. I used to be able to factor certain trinomials of the 5th degree using carefully placed zero. Now I'm trying to remember which ones and how...

 

[pwsnafu]

Yes! I am pleasantly surprised at how quickly you factored x^4 + x^2 + 1

This is a thread about math tricks (see title) and i know a trick to factoring x^4 + x^2 + 1

This is how I did it: I know that there is no linear factor, so it's quadratic times quadratic. The coefficient of X⁴ is 1, and the constant term is 1. So one of the factors is X² + a X + 1. And just try different a. Naturally you try a=1 first. And you are done.

While we are on the topic of "no complex number tricks", when a student first learns complex numbers I give them this little gem:

Suppose x + y = 2 and xy = 3. Find 1/x + 1/y.

Nine times out ten, the student will try to use the quadratic formula. If the student doesn't know about complex numbers they'll stop and try to find a different method. The students that do know complex numbers will just keep going.

 

[micromass]

micromass asked...

"Well, how would you solve $x^2=2+i$?? By saying $x=\sqrt{2+i}$?? How helpful..."

I would throw caution to the wind and let the chips fall where they may...THEN...

I would put this particular example into the quadratic formula and then sustitute back to confirm i got the right answer. Messere Gauss proved I can do it this way.

By all means, try it. You'll end up with $x=\sqrt{2+i}$. But that doesn't really help if you want a solution in the form $a+bi$

If you post a more complicated example which does not have a closed form solution i can still get close to the answer by using the 4 binary operations AND THE EXTRACTION OF ROOTS.

The ability to extract roots of real numbers is necessary, this means you need to extract roots of negative values.

The square root symbol has nothing to do with polynomials, really nothing. The reason why we can factor polynomials is because we introduced a number i such that $i^2=-1$. Wheter we write $\sqrt{-1}=i$ is irrelevant.

The Real numbers are 1 dimensional, Complex numbers NOT on the vertical axis have 2 distinct dimensions.

How fortunate that i is on the vertical axis and has length 1? So we can use it as a 'cheat' to simplify the algebra.

You are correct as far as the cyclotomic x^n = a + bi is concerned

However using sines and cosines to get the answer of course produces multi values since they are periodic!

No, I only get n values.

This is a neat trick, and i have solved x^n = a + bi in Linear Algebra so i know what you mean and agree with you on most of your points.

micromass also said...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of $x^n=-1$:

To the best of my knowledge this method you describe works only if you don't have a MIDDLE TERM.

x^2 = (1 + 2i)x + i would not work with this method but easily works using quadratic formula.

Yes, using the quadratic formula is fine. But using the square root symbol is not necessary in the quadratic formula!

 

[agentredlum]

$${(}{{i}^{4}{)}}^{1/2}{=}{(}{\sqrt{i}{)}}^{4}$$

You agree to this[Your rule does not suggest this unless you incorporate some exception handling clause]?

What about [(-4)^(1/2)]^6 = [(-4)^6]^(1/2)]

The left is definitely-64 if you wish to avoid i, multiply the exponents.The right is not definitely -64 because you can simplify inside the brackets without fear of complex. As a matter of fact the right is +64 by convention but the left is -64 no matter how you evaluate it, so exchanging exponents like that is dangerous even for real negative numbers, let alone i and -i

 

[agentredlum]

This is how I did it: I know that there is no linear factor, so it's quadratic times quadratic. The coefficient of X⁴ is 1, and the constant term is 1. So one of the factors is X² + a X + 1. And just try different a. Naturally you try a=1 first. And you are done.

While we are on the topic of "no complex number tricks", when a student first learns complex numbers I give them this little gem:

Suppose x + y = 2 and xy = 3. Find 1/x + 1/y.

Nine times out ten, the student will try to use the quadratic formula. If the student doesn't know about complex numbers they'll stop and try to find a different method. The students that do know complex numbers will just keep going.

I don't need quadratic formula or complex numbers for this, the answer is 2/3

I did it in my head in 5 billion nano-seconds

 

[agentredlum]

By all means, try it. You'll end up with $x=\sqrt{2+i}$. But that doesn't really help if you want a solution in the form $a+bi$
The square root symbol has nothing to do with polynomials, really nothing. The reason why we can factor polynomials is because we introduced a number i such that $i^2=-1$. Wheter we write $\sqrt{-1}=i$ is irrelevant.Yes, using the quadratic formula is fine. But using the square root symbol is not necessary in the quadratic formula!

Yes you are correct i suppose it's like solving x^2 = 4 and expecting a different result by using quadratic formula. That was silly of me, i should have tried it first. However that is not what i considered my important point, and I've been up all night and i can't remember what my point was/is anymore.

Please show me this quadratic formula you speak of that has no radicals.

 

[micromass]

Yes you are correct i suppose it's like solving x^2 = 4 and expecting a different result by using quadratic formula. That was silly of me, i should have tried it first. However that is not what i considered my important point, and I've been up all night and i can't remember what my point was/is anymore.

Please show me this quadratic formula you speak of that has no radicals.

Let $ax^2+bx+c=0$ be your equation. Let $D=b^2-4ac$, consider $Z$ such that $Z^2=D$, then the solutions are

$$x=\frac{-b\pm Z}{2a}$$

See, no radicals involved, but you still have the same formula This shows that you don't need the square root symbol anywhere...

 

[agentredlum]

Let $ax^2+bx+c=0$ be your equation. Let $D=b^2-4ac$, consider $Z$ such that $Z^2=D$, then the solutions are

$$x=\frac{-b\pm Z}{2a}$$

See, no radicals involved, but you still have the same formula This shows that you don't need the square root symbol anywhere...

Please post picture, I CAN'T READ IT!

it comes out as /frac-b/pm Z gibberish in my browser